A particular variety of watermelon weighs on average 21.3 pounds with a standard

Question

A particular variety of watermelon weighs on average 21.3 pounds with a standard deviation of 1.07 pounds. Consider the sample mean weight of 9 watermelons of this variety. Assume the individual watermelon weights are independent.

a. What is the expected value of the sample mean weight? Give an exact answer.  

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.  

c. What is the approximate probability the sample mean weight will be greater than 21.3? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.  

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.93? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.

Explanation / Answer

We will use the t-distribution to solve the problem because the sample size is just 9 , which is less than 30 ( the min required for using Z distribution)

Sample mean = pop. mean = 21.3

Sample deviation = pop. deviation = 1.07/sqrt(9) = .3567

a. The expected value of sample weight = sample mean = 21.3

b. Sample deviation = (1.07/sqrt(9)) = .3567

c. P(X21.3) = P(t> 21.3-21.3/(1.07/sqrt(9)) ) = P(t>0) = .50

d. P(X<c) = .93

We get the t score for df of n-1 = 8, 1.4450

So, c= Mean + t*Sigma/sqrt(n) = 21.3+/- 1.4450*1.07/sqrt(9) = 21.8154

So, value of c such that .93 is below it is 21.8154

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